A) Iron and \[{{H}_{2}}S{{O}_{4}}\] (aqueous)
B) Iron and steam
C) Copper and \[HCl\] (aqueous)
D) Sodium and ethyl alcohol
Correct Answer: C
Solution :
| [c] \[Fe+dil\text{ }{{H}_{2}}S{{O}_{4}}\to FeS{{O}_{4}}+{{H}_{2}}\,\uparrow \] |
| \[3Fe+\underset{Steam}{\mathop{4{{H}_{2}}O}}\,\to ~F{{e}_{3}}{{O}_{4}}+4{{H}_{2}}\uparrow \] |
| Cu is below H in electrochemical series. |
| \[Cu+dil.\,HCl\xrightarrow[{}]{{}}No\,\,\,reaction\] |
| \[2Na+{{C}_{2}}{{H}_{5}}OH\xrightarrow[{}]{{}}2{{C}_{2}}{{H}_{5}}ONa+{{H}_{2}}\uparrow \] |
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